Question

Which of the following functions form Z to itself are bijections?
(a) $f\left(x\right)=x^3$
(b) $f\left(x\right)=x+2$
(c) $f\left(x\right)=2x+1$
(d) $f\left(x\right)=x^2+x$

Answer 1

$$\begin{gathered} \left(a\right)f\text{is not onto because for}\mathit{\text{y}}\text{= 3∈Co-domain(}\mathit{\text{Z}}\text{), there is no value of x∈Domain(Z)} \\ {x}^3=3 \\ \Rightarrow x=\sqrt[3]{3}\notin Z \\ \text{⇒}f\text{is not onto.} \\ \text{So,}f\text{is not a bijection.} \end{gathered}$$

(b) Injectivity:
Let x and y be two elements of the domain (Z), such that
$$\begin{gathered} x+2=y+2 \\ \Rightarrow x=y \end{gathered}$$
So, f is one-one.

Surjectivity:
Let y be an element in the co-domain (Z), such that
$$\begin{gathered} y=f\left(x\right) \\ \Rightarrow y=x+2 \\ \Rightarrow x=y-2\in Z\left(\text{Domain}\right) \end{gathered}$$
$\Rightarrow$f is onto.
So, f is a bijection.

$$\begin{gathered} \left(c\right)f\left(x\right)=2x+1\text{is not onto because if we take 4 ∈}Z\left(\text{co domain}\right),\text{then}4=f\left(x\right) \\ \Rightarrow 4=2x+1 \\ \Rightarrow 2x=3 \\ \Rightarrow x=\frac{3}{2}\notin Z \\ \text{So,}f\text{is not a bijection.} \\ \left(d\right)f\left(0\right)=0^2+0=0 \\ \mathrm{and}f\left(-1\right)={\left(-1\right)}^2+\left(-1\right)=1-1=0 \\ \text{⇒0 and -1 have the same image.} \\ \text{⇒}f\text{is not one-one.} \\ \text{So,}\mathit{\text{f}}\text{is not a bijection.} \end{gathered}$$

So, the answer is (b).